Critical phenomena in gravitational collapse ofHusain-Martinez-Nunez scalar field

Xiaobao Wang∗1, Xiaoning Wu†2, and Sijie Gao‡3

1Department of Physics, Beijing Normal University, Beijing , China,1008752Institute of Applied Mathematics, Academy of Mathematics and System Science,

Chinese Academy of Sciences, P.O. Box 2734, Beijing, China, 100080

April 30, 2019

Abstract

We construct analytical models to study the critical phenomena in gravitational collapse ofthe Husain-Martinez-Nunez massless scalar field. We first use the cut-and-paste technique tomatch the conformally flat solution (c = 0 ) onto an outgoing Vaidya solution. To guaranteethe continuity of the metric and the extrinsic curvature, we prove that the two solutions mustbe joined at a null hypersurface and the metric function in Vaidya spacetime must satisfysome constraints. We find that the mass of the black hole in the resulting spacetime takes theform M ∝ (p−p∗)γ , where the critical exponent γ is equal to 0.5. For the case c 6= 0, we showthat the scalar field must be joined onto two pieces of Vaidya spacetimes to avoid a nakedsingularity. We also derive the power-law mass formula with γ = 0.5. Compared with previousanalytical models constructed from a different scalar field with continuous self-similarity, weobtain the same value of γ. However, we show that the solution with c 6= 0 is not self-similar.Therefore, we provide a rare example that a scalar field without self-similarity also possessesthe features of critical collapse.

1 IntroductionGravitational collapse is the main reason of various galactic structures and it remains one

of the most interesting and fundamental problems in general relativity. The end state ofgravitational collapse could be a black hole, naked singularity or flat spacetime. In a seminalwork by Choptuik [1], some intriguing and universal properties concerning the formation ofblack holes from massless scalar fields were found. This is called the critical phenomenon.Particularly, near the threshold, the black hole mass can always be expressed in the form ofpower-law:

M ∝ (p− p∗)γ , (1)

for p > p∗, where p is a parameter of the initial data to the threshold of black hole formation.Numerical simulations have shown that that the critical exponent γ is equal to 0.5 for solutionswith continuous self-similarity (CSS) and γ ≈ 0.37 for solutions with discrete self-similarity(DSS). Details about the critical phenomenon can be learned in [2, 3].

In addition to numerical calculation, analytical models were also built to explore the criticalphenomena. Patrick R. Brady [4] studied an exact one parameter family of scalar field solutionswhich exhibit critical behaviours when black hole forms. J. Soda and K. Hirata [5] analyticallystudied the collapse of continuous self-similar scalar field in higher dimensional spacetimes andfound a general formula for the critical exponents which agrees with the exponent γ = 0.5for n = 4. A. Wang et al [6] constructed an analytical model by pasting the BONT model(a massless scalar field) with the Vaidya model. They demonstrated that the black hole massobeys the power law with γ = 0.5. A. Wang et al [7] also analytically studied the gravitationalcollapse of a massless scalar field with conformal flatness. They showed that the mass of the

∗bao@mail.bnu.edu.cn†wuxn@amss.ac.cn‡Corresponding author: sijie@bnu.edu.cn

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black hole without self-similarity turns on with finite nonzero values. Recent developmentsregarding the critical phenomenon can be found in Refs. [8–14].

In this paper, we investigate the critical phenomena associated with an exact scalar fieldsolution discovered by Husain-Martinez-Nunez (HMN) [15]. The HMN solution is interestingbecause it represents a black hole in a FLRW universe [16–18]. It was pointed out that theHMN solution brings in new phenomenology(S-curve) of apparent horizon [19]. Moreover, theconformally transformed HMN spacetime can be a inhomogeneous vacuum solution in Brans-Dick theory [20–22]. Following the treatment in [6], we match the HMN solution onto anoutgoing Vaidya solution along a null hypersurface. This is to guarantee that the black holemass is finite. Usually, the hypersurface connecting the two parts of the spacetime is a thinshell, i.e., the extrinsic curvature across the hypersurface is discontinuous. By applying theDarmois-Israel formula [28], one can find the relationship between the jump of the extrinsiccurvature and the surface stress-energy tensor of the thin shell. However, by properly choosingthe function m(U) in the Vaidya metric, the extrinsic metric can be continuous across the nullsurface. Therefore, no thin shell forms and the resulting spacetime can be at least C1.

After the matching, we calculate the apparent horizon and define the black hole mass asthe Komar mass at the intersection of the apparent horizon and the null hypersurface. Wefirst study the case c = 0 where the HMN solution is conformally flat and has CSS. The massof the black hole is found in the power-law form M ∝

√−a for a < 0, which means γ = 0.5.

When a = 0, the black hole disappear and the spacetime becomes Minkowski.The case of c 6= 0 is more complicated. This solution has no self-similarity. We still find

M ∝√−a for a < 0. Differing from the case of c = 0, we show that the limiting spacetime

(a = 0) possesses a naked singularity with non-zero ADM mass. Therefore, there is a massgap between the black hole spacetime (a < 0) and the spacetime with naked singularity a = 0.

The paper is organized as follows. In section 2, we briefly introduce the Husain-Martinez-Nunez (HMN) scalar field solution. In section 3, by using the cut-and-paste method, we matchthe conformally flat HMN solution (c = 0 with CSS) onto an outgoing Vaidya spacetime atboth timelike and null hypersurfaces to construct an analytical model. It turns out thatonly the matching along the null hypersurface can guarantee the continuity of the metric andextrinsic curvature across the surface. Then, we use this analytical model to study the criticalphenomenon and derive the mass formula. In section 4, we join a general HMN (a 6= 0, c 6= 0)with two outgoing Vaidya spacetimes. We show that the mass of the black hole approacheszero for a < 0. We also find that the critical spacetime (a = 0) possesses a naked singularitywith nonzero ADM mass. Concluding remarks are given in Section 5. In Appendix A, weprove that the HMN solution has CSS only when and a 6= 0 and c = 0.

2 Husain-Martinez-Nunez (HMN) spacetimeThe Husain-Martinez-Nunez spacetime [15] satisfies the Einstein-scalar field equations

Gab = 8πTab , (2)

Tab = ∇aΦ∇bΦ−1

2gabg

cd∇cΦ∇dΦ . (3)

The spherically symmetric solution is given by 1

ds2 = (at+ 1)

[−(

1− 2c

r

)αdt2 +

(1− 2c

r

)−αdr2 + r2

(1− 2c

r

)1−α

dΩ2

], (4)

Φ(r, t) = ± 1

4πln

[(1− 2c

r

) α√3

(at+ 1)√

3

], (5)

where α = ±√

32. From the Ricci scalar

R =12ca2(r − c)− 3a2r2

2r2(at+ 1)3

(1− 2c

r

)−2−α

+2c2(1− α2)

(at+ 1)r4

(1− 2c

r

)−2+α

, (6)

we see that the curvature singularities are located at r = 2c (timelike singularity) and t =−1/a(spacelike singularity). Using Θ± to label the expansion of null geodesics, we have

Θ+ =1√h

∂√h

∂λ+, (7)

Θ− =1√h

∂√h

∂λ−, (8)

1 Without loss of generality, we have set b = 1 in the original metric in [15].

2

where√h = R2 sin θ = (at+ 1)r2

(1− 2c

r

)1−αsin θ, and λ± is the affine parameter of the null

geodesics. The tangent to the null geodesic with affine parameter is

∂

∂λ+=

1

(at+ 1)

(1− 2c

r

)−α∂

∂t+

1

(at+ 1)

∂

∂r, (9)

∂

∂λ−=

1

(at+ 1)

(1− 2c

r

)−α∂

∂t− 1

(at+ 1)

∂

∂r. (10)

We can get

Θ+ =

ar2

(1− 2c

r

)−α+

[2r − 2cr(α− 1)

r − 2c

](at+ 1)

r−2(at+ 1)−2 , (11)

Θ− =

ar2

(1− 2c

r

)−α−[2r − 2cr(α− 1)

r − 2c

](at+ 1)

r−2(at+ 1)−2 . (12)

The apparent horizon satisfies Θ+ = 0,Θ− < 0, which is located at

a

atAH + 1= − 2

r2AH

[rAH − c(1 + α)] (1− 2c

rAH)α−1 . (13)

The detailed analyses about the HMN sapcetime can be found in [18]. We shall focus on thecase −∞ ≤ t ≤ − 1

aand a < 0 because it corresponds to a black hole solution. When a 6= 0

and α =√

32, the apparent horizon in the HMN spacetime has "S-curve" shape. However,

this does not make differences in our results. Therefore, in this paper, we just study the caseα = −

√3

2for simplicity. The results remain true when α =

√3

2. The MisnerSharp mass [34] is

defined by

M =R

2

(1− gab∇aR∇bR

). (14)

whereR denotes the areal radius. From [19], we know that in spherically symmetric spacetimes,the apparent horizon satisfies

gab∇aR∇bR = 0 . (15)

Therefore, on the apparent horizon, the Misner-sharp mass becomes

M =R

2. (16)

3 Critical behaviour of HMN scalar field with confor-mal flatness (c 6= 0)

3.1 Matching at a timelike boundaryTo study the critical phenomenon of HMN massless scalar field, we start with the simple

case c = 0, where the spacetime is conformally flat [15]. First, we need to join the HMNsolution with an outgoing Vaidya solution such that the resulting spacetime is asymptoticallyflat. In this section, we assume that the boundary connecting the two solutions is a timelikehypersurface. We shall use "−" to label the inner HMN spacetime and "+" to label theexterior Vaidya spacetime (see Fig. 1).

Figure 1: a 6= 0, c = 0 timelike hypersurface

3

The interior spacetime is described by Eq. (4) for c = 0

ds2− = (1 + at)

[−dt2 + dr2 + r2 (dθ2 + sin θ2dφ2)] . (17)

The exterior spacetime is described by the outgoing Vaidya metric

ds2+ = −f+dU

2 − 2dUdR+R2dΩ2 , (18)

where f+ = 1 − 2m(U)R

. We choose ξi = λ, θ, φ as the intrinsic coordinates on the hyper-surface. Σ is determined by functions t(λ), r(λ) from the interior and T (λ), R(λ) from theexterior. The induced metric on Σ is given by

ds2−

∣∣∣Σ

= (at+ 1)[(−t2 + r2)dλ2 + r2 (dθ2 + sin θ2dφ2)] , (19)

ds2+

∣∣∣Σ

= −(f+U + 2R

)Udλ2 +R2dΩ2 . (20)

Here the do“.” means the derivative with respect to λ.We use the Darmois junction conditions to match the solutions across Σ

ds2−

∣∣∣Σ

= ds2+

∣∣∣Σ, (21)

k−ab

∣∣∣Σ

= k+ab

∣∣∣Σ, (22)

where kab is the extrinsic curvature of Σ.Denote the coordinates of the four-dimensional spacetime by xµ. Σ is determined by the

functions xµ(ξi). Then the components of kab can be calculated from

kij = −nµ∂2xµ

∂ξiξj− nνΓνµρ

∂xµ

∂ξi∂xρ

∂ξj(23)

where na is the spacelike normal to Σ. Computing kij from the interior and exterior, respec-tively, we obtain the nonvanishing components

k−λλ =ar(r2 − t2)− 2(at+ 1)(tr − rt)

2√

(at+ 1)(t2 − r2), (24)

k+λλ =

U(f+U2 + 2RU

) 12

[− U

2

R

(mf+

R− dm

dU

)+ R

(U

U− 3

Um

R2

)− R

], (25)

k−θθ =1

sin2 θkIφφ=

ar2r + 2(at+ 1)tr

2√

(at+ 1)(t2 − r2), (26)

k+θθ =

1

sin2 θkEφφ=

R(f+U2 + 2RU

) 12

(R+ f+U

). (27)

Substituting Eqs. (19) and (20) into Eq. (21), we have

R =√at+ 1r , (28)

(at+ 1)(t2 − r2) =(f+U + 2R

)U . (29)

Substituting Eqs. (26) and (27) into Eq. (22), with the help of Eq. (29), we obtain

f+U + R =arr + 2(at+ 1)t

2√at+ 1

. (30)

Eq. (28) yields

R =art

2√

(at+ 1)+√at+ 1r . (31)

Therefore, one can solve Eqs. (29)-(31) and obtains

U =2(at+ 1)

32(t− r

)ar + 2(at+ 1)

. (32)

f+ = 1− m

R= 1− a2r2

4(at+ 1)2. (33)

4

Thus,

m =a2r3

8(at+ 1)32

. (34)

To proceed, we calculate the following derivatives:

U =

√at

(r + 2t)2

[2tr2 − (4t+ 3r)tr + 2t(2tt− 2tr + t2) + r(2tt− 2tr + 3t2)

], (35)

R =√atr +

2√attr +

√a(2ttr + 2trt− rt2)

4t3/2, (36)

dm(U)

dU=

m

U=

3r2(r + 2t)(rt− 2tr)

32t4(r − t). (37)

where t = t + a−1. Substitute the above results into Eq. (25) and according to Eq. (22),letthe right-hand side of Eq. (24) be equal to the right-hand side of Eq. (25). After a lengthycalculation, we obtain the following result, which is surprisingly simple

(t− r)(t+ r)2 = 0 (38)

Obviously, the solution is r = t or r = −t. But this means that the hypersurface is null,inconsistent with our assumption. Therefore, we conclude that the two spacetimes cannot bematched through a timelike hypersurface if the continuity of the extrinsic curvature is required.

3.2 Matching at a null hypersurfaceMatching the two solutions at a null hypersurface is more complicated than at a time-

like hypersurface. We shall follow the method in [6] and [26]. First we use the coordinatetransformation v = t + r to replace the coordinate r in Eq. (4) and obtain the metric in theinterior

ds2 = − [a (v − r) + 1] dv(dv − 2dr) +R2 (dθ2 + sin2 θdφ2) . (39)

where

R2 = [a(v − r) + 1] r2 . (40)

Let Σ be the null hypersurface v = v0. The normal to Σ is

n−a = s−1dva , (41)

where s is a negative arbitrary function such that na− is a future directed vector. We canintroduce a transverse null vector Na by requiring

naNa = −1 , (42)

NaNa = 0 . (43)

Without loss of generality, we assume that N−a = Nvdva + Nrdra. Then it is easy to showthat

N−a = s[a (v − r) + 1]

(1

2dva − dra

). (44)

Now we choose s = − 1

a(v − r) + 1

∂R

∂r. Choose ξi = R, θ, φ to be the intrinsic coordinates

on Σ. Then Σ can be determined by

v = v0 , r = r(R) , θ = θ , φ = φ , (45)

where r(R) is determined by Eq. (40) with v = v0. We define e−µ(a) ≡∂xµ−∂ξa

as given in the [6].

Thus

e−r(1) =2rR

2R2 − ar3, e−θ(2) = 1 , e−φ(3) = 1 . (46)

Similarly to Eq. (23), the transverse extrinsic curvature for the null surface is given by [26].

kij = −Nµ∂2xµ

∂ξiξj−NνΓνµρ

∂xµ

∂ξi∂xρ

∂ξj(47)

5

Figure 2: ingoing null hypersurface v − v0 = 0 or U − U0(R) = 0

By straightforward calculation, we find

k−ab =−3a2r

(3ar − 2av0 − 2)2√a (v0 − r) + 1

dRadRb+r [a (r − 2v0)− 2] (3ar − 2av0 − 2)

8 [a (v0 − r) + 1]32

(dθadθb + sin θ2dφadφb

).

(48)On the other hand, we need to join the HMN solution to the exterior Vaidya spacetime

(18) at Σ, as shown in Fig. 2. Assume that the null surface Σ can be described by U = U0(R)from the Vaidya solution. It follows from (18) that

dU0

dR= − 2R

R− 2m[U0(R)]. (49)

The spacetime coordinates xµ+ can be expressed as functions of ξi:

U = U0(R) , R = R , θ = θ , φ = φ . (50)

Define e+µ(i) ≡

∂xµ+∂ξi

and it is easy to find

e+U(1) = − 2

f+, e+R

(1) = e+θ(2) = e+φ

(3) = 1 . (51)

The normal to Σ is given by

n+a = β−1∇a(U − U0(R)) = β−1

(dUa +

2

f+dRa

). (52)

where β is a negative function which will be determined later. Then the transverse null vectorNa in Eq. (42) is

N+a =

βf+

2dUa , (53)

The continuity condition on Σ requires [26]

N+µ e

+µ(i)

∣∣∣v=v0

= N−µ e−µ(i)

∣∣∣v=v0

. (54)

This also guarantees that the normal vectors na defined on both sides are the the same.According to Eqs. (44),(46)-(46), Eqs. (51), (53), we find that the nontrivial equations in Eq.(54) are

N−µ e−µ(1)

∣∣∣v=v0

= N−r e−r(1) = 1 , (55)

N+µ e

+µ(1)

∣∣∣v=v0

= N+U e

+U(1) = −β . (56)

Hence, we get β = −1.Now we can calculate the corresponding transverse extrinsic curvature from Eqs. (47) and

obtain

k+ab =

−2m′(r)r′(R)√a (v0 − r) + 1r − 2m(r)

dRadRb +

(R

2−m[U0(R)]

)(dθadθb + sin2 θdφadφb

). (57)

Since k+ab = k−ab, Eqs. (48) and (57) give rise to

k+RR

∣∣Σ

= k−RR∣∣Σ. (58)

6

By integration, we obtain m(r) as

m(r) =−8 + a3

(−27r3 + 12rv2

0 − 8v30

)+ 12a [r − 2(v0 + 18c1)] + 24a2 [−v(v0 + 18c1) + r(v0 + 27c1)]

−216a [1 + a(v0 − r)]32

, (59)

where c1 is an integral constant. Using k+θθ

∣∣Σ

= k−θθ∣∣Σ, we can fix c1:

c1 =−(av0 + 1)2

54a. (60)

Consequently,

m(r) =a2r3

8 [a (v0 − r) + 1]3/2. (61)

Note thatm(r)

∣∣Σ

= m(U(R(r))∣∣Σ. (62)

So, Eq. (61) together with Eq. (49) specifies a unique metric function m(U) in the Vaidyasolution. Therefore, we have matched the conformally flat spacetime with the Vaidya spacetimeat the null hypersurface.

3.3 Mass of the black holeFrom Eq. (14), one can calculate the Misner-Sharp mass for the HMN spacetime described

by metric (17) and obtain

M =r3a2

8 (at+ 1)32

. (63)

Therefore, m(r) in Eq. (61) is just the Misner-Sharp mass at v = v0. The apparent horizondetermined by Eq. (13) takes the simple form for c = 0:

a

atAH + 1= − 2

rAH. (64)

Note that the null surface is determined by

v0 = t+ r , (65)

Eqs. (64) and (65) immediately gives the coordinates at the intersection of Σ and the apparenthorizon:

ri = 2v0 +2

a, (66)

ti = −v0 −2

a. (67)

Since r > 0 and a < 0, from Eq. (66), we see that the existence of the intersection requires

v0 > |a|−1 (68)

Therefore, the Misner-Sharp mass at the intersection is

Mi =√−a(v0 +

1

a

) 32

. (69)

As is known, the event horizon coincides with the apparent horizon in the outgoing Vaidyaspacetime as shown in Fig. 3. It is also known that the mass function m in the Vaidya metricis constant alone the event horizon [35]. Thus, it is natural to take the mass in Eq. (69) to bethe mass of the black hole.

To investigate the critical behavior as a→ 0, we impose the condition that ri in Eq. (67)does not change with a. This means that v0 must take the form

v0 = V − 1

a(70)

where V is a positive constant independent of a. Thus, Eq. (69) gives the mass of black hole:

Mbh =√−aV

32 . (71)

Eq. (71) shows that the mass of black hole can be put in the form of Eq. (1) and the scalingexponent is γ = 0.5. Obviously, as a approaches zero, the mass of the black hole vanishes andthe spacetime becomes Minkowski.

7

Figure 3: Penrose diagram for HMN spacetime (a 6= 0, c = 0) matching with an outging Vaidyaspacetime. The singularity is located at t = −a−1. The HMN spacetime matches with the outgoingVaidya spacetime at v = v0.

4 Collapse of the general HMN scalar fieldIn this section, we shall investigate the gravitational collapse associated with a general

HMN scalar field (a 6= 0, c 6= 0).

4.1 Matching to an outgoing Vaidya solution at a null hyper-surface

Under the coordinate transformation

v = t+ h(r) (72)

Eq. (4) can be rewritten in the form

ds2 = − [a(v − h(r)) + 1]

(1− 2c

r

)α (dv2 − 2h′(r)dvdr

)+a [(v − h(r)) + 1]

(1− 2c

r

)1−α

r2dΩ2 .

(73)Here, the function h(r) satisfies

h′(r) =

(1− 2c

r

)−α. (74)

The areal radius R takes the form

R =√a(v − h(r)) + 1

(1− 2c

r

) 1−α2

r . (75)

Similarly to section 3, we match the solution with an outgoing Vaidya solution at the nullhypersurface v = v0 (see Fig. 4). Substitution of v = v0 into Eq. (75) yields the functionr = r(R). By the method in section 3.2, the extrinsic curvature can be calculated as

k−ab =

(ah′(r)r′(R)

1 + a(v0 − h(r))− r′′(R)

r′(R)

)dRadRb −

[(2 +

√3)c− 2r + 4ac(1− 2c/r)−

√3

2 r − 2a(1− 2c/r)−√

32 r2

4 [1 + a(v0 − h(r))] r′(R)

+((2 +

√3)c− 2r)ah(r) + 2arv0 + ar(2c− r)h′(r)− (2 +

√3)acv0

4 [1 + a(v0 − h(r))] r′(R)

] (dθadθb + sin2 θdφadφb

).

(76)

8

where

r′(R) = −2(1− 2c

r)α−12 (r − 2c)

√a [v0 − h(r)](

(2 +√

3)c− 2r)

(1 + av0)− a((2 +√

3)c− 2r)h(r) + ar(−2c+ r)h′(r), (77)

r′′(R) = −[((2 +

√3)c− 2r)(1 + av0)− a((2 +

√3)c− 2r)h(r) + ar(−2c+ r)h′(r)3

]−3

+

2(1− 2c/r)√

32[a2c2h(r)2 + 2a((2 +

√3)c− 2r)(2c− r)r(1 + av0)h′(r) + a2r2(−2c+ r)2h′(r)2

− 2ah(r)(c2(1 + av0) + a((2 +√

3)c− 2r)(2c− r)rh′(r) + ar2(−2c+ r)2h′′(r))

+ (1 + av0)(c2(1 + av0) + 2ar2(−2c+ r)2h′′(r))]. (78)

It is easy to see that k+ab takes the same form as Eq. (57). Then k−ab = k+

ab yields

m(r) =1

2(1− 2c/r)(2−

√3)/4r

√a(v0 − h(r)) + 1 +

1

4(a(v0 − h(r)) + 1)r′(R)

((2 +

√3)c− 2r

+ 4acr(1− 2c/r)−√

32 − 2a(1− 2c/r)−

√3

2 r2 + ((2 +√

3)c− 2r)av0

− a((2 +√

3)c− 2r)h(r) + ar(−2c+ r)h′(r)).

Thus, by our construction, the metric of the resulting spacetime is continuous and the extrinsiccurvature of the null hypersurface is also continuous. Therefore, we have shown that the generalHMN spacetime ( a 6= 0, c 6= 0) and Vaidya spacetime can be matched at a null hypersurfaceas showed in Fig. 4.

Figure 4: Penrose diagram for the HMN solution(a 6= 0, c 6= 0, α = −√32 ) matching with an

outging Vaidya solution. There are two singularities at t = | 1a | and r = 2c, where r = 2c is a nakedsingularity.

4.2 Mass of the black holeBy the argument in section 3.3, one can show that m(r) in Eq. (79) is exactly the Misner-

sharp mass for the metric in Eq. (73). Therefore, from Eqs. (13),(16) and (75), we can obtainthe mass on the apparent horizon mAH(r):

mAH(r) =RAH(r)

2=√−ar2[8r − 8c(1 + α)]−

12

(1− 2c

r

)−α+1

. (79)

The ingoing null hypersurface boundary is defined by

v0 = t+ h(r) . (80)

From the Eq. (13) and the Eq. (80), we can get the coordinates (ri, ti) at the intersection ofthe apparent horizon and the null hypersurface v = v0, which satisfies

9

v0 +1

a=

(1− 2c/ri)√

32 (2c− ri)ri +

[(−2 +

√3)c+ 2ri

]h(ri)

(−2 +√

3)c+ 2ri(81)

ati + 1 =a(1− 2c/ri)

√3

2 ri(2c− ri)−2c+

√3c+ 2ri

> 0 . (82)

Similarly, we choose the null hypersurface which intersects with the apparent horizon at afixed radius ri = r0, i.e., independent of a. Again, we take the Misner-Sharp mass at theintersection as the black hole mass. Then, Eq. (79) gives the mass of the black hole

Mbh(r0) =√−af(r0, c) , (83)

where

f(r0, c) = r20[8r0 − 8c(1 + α)]−

12

(1− 2c

r0

)−α+1

. (84)

Eq. (83) shows clearly that the black hole mass satisfies the power law with γ = 0.5. However,the spacetime for a < 0 possesses a naked singularity r = 2c (see Fig. 4), in violation ofthe cosmic censorship conjecture. To remove the naked singularity, we join another outgoingVaidya spacetime at v = v1(v1 < v0), as shown in Fig. 5. No naked singularity exists in thisnew spacetime.

When we study the relation between the mass and the parameter a, we treat c as a constant.We see that Mbh → 0 as a → 0. When a = 0, there is no black hole but a naked singularityas shown in section 4.3.

Figure 5: Penrose diagram for the HMN scalar field (a 6= 0, c 6= 0) matching with two outgoingVaidya spacetimes. We see that the naked singularity in Fig. 4 has been replaced by the Vaidyaspacetime.

4.3 "Critical spacetime": a = 0 and c 6= 0

Now we study the critical HMN spacetime. For a = 0, Eq. (4) becomes

ds2 = −(

1− 2c

r

)αdt2 +

(1− 2c

r

)−αdr2 + r2

(1− 2c

r

)1−α

dΩ2 , (85)

To calculate the apparent horizon of the spacetime, we first choose two families of radial nullvector fields

∂

∂λ+=

(1− 2c

r

)−α∂

∂t+

∂

∂r(86)

∂

∂λ−=

(1− 2c

r

)−α∂

∂t− ∂

∂r. (87)

10

where λ± is the affine parameter of the null geodesic.According to Eq. (11)

Θ+ = −2 (c− r + cα)

r(−2c+ r), (88)

Θ− =2(c− r + cα)

r(−2c+ r). (89)

Therefore, the spacetime has no apparent horizon when r > 2c. According to Eq. (6), thespacetime possesses a naked singularity at r = 2c (see Fig. 6).

Figure 6: Penrose diagram for the HMN spacetime(a = 0, c 6= 0). There is a naked singularity atr = 2c.

Since the spacetime is asymptotically flat, we calculate the ADM mass and find

M = cα . (90)

When a 6= 0, the mass of the black hole is given by Eq. (83), which shows clearly that M → 0as a → 0. However, when a = 0, as we just discussed, the spacetime is not Minkowski andits ADM mass is nonzero. Therefore, there exists a mass gap between the black hole solution(a 6= 0) and its limiting spacetime (a = 0).

5 ConclusionIn this paper, we have used the "cut and paste" method to construct analytical models

and study the critical phenomena of the HMN scalar filed. We have shown that the HMNsolution with conformal flatness (c 6= 0) can be matched with the Vaidya solution along a nullhypersurface, but not a timelike hypersurface. We have derived the differential equation whichspecifies the metric function in the Vaidya solution. For c 6= 0, we have joined the scalar fieldonto two pieces of Vaidya spacetimes to avoid the naked singularity.

We have studied the gravitational collapse for the HMN scalar field and shown that blackhole mass satisfies the power law with γ = 0.5. This is consistent with previous results in theliterature. When c 6= 0, the HMN spacetime has no CSS and the black hole also turns onat infinitely small mass. The result is different from the model in [7], which shows that theformation of black holes may turn on at finite mass when the gravitational collapse has noself-similarity. On the other hand, the mass gap exists between the black hole and the nakedsingularity during the gravitational collapse of HMN scalar field when c 6= 0 as discussed insection 4. Our work suggests that critical collapse can be studied from analytical models whichare constructed by known solutions. More models should be investigated in the future in orderto test the universal features in gravitational collapse.

A Self-similarity of HMN spacetimeIn this appendix, we will prove that the HMN spacetime is CSS (continuous self similar)

only when c = 0 and a 6= 0. A spacetime is continuous self-similar if there exists a conformal

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Killing vector field ξa satisfying∇(aξb) = gab . (91)

As a result of the spherical symmetry, we can write

ξa = x

(∂

∂t

)a+ y

(∂

∂r

)a. (92)

Here, x and y are functions of r and t. Substituting this expression into Eq. (91), we find that

yR,r + xR,t = R , (93)yν,r + xν,t + y,r = 1 , (94)yλ,r + xλ,t + x,t = 1 , (95)my,t − nx,r = 0 . (96)

Here,

R2 = (at+ 1) r2

(1− 2c

r

)1−α

,

λ =1

2log

[(at+ 1)

(1− 2c

r

)α],

ν =1

2log

[(at+ 1)

(1− 2c

r

)−α],

m = (at+ 1)

(1− 2c

r

)−α,

n = (at+ 1)

(1− 2c

r

)α. (97)

From Eq. (93), we can get

x = − 1

ar(−2c+ r)(1 + at)(4cr − 2r2 − 2cy +

√3cy + 2ry) . (98)

Substituting Eq. (98) into Eq. (94), we obtain

y =√r√−2c+ rD(t) , (99)

where D(t) is a integration function of t. Putting Eqs. (98) and (99) into Eq. (96), we find

−c(1− 2c

r)−√

3((−2 +√

3)c−√

3r)

a(−2c+ r)2r2= − D′(t)

(1 + at)D(t)≡ C0 . (100)

Obviously, C0 must be a constant independent of r and t. So the only solution is

c = 0 , (101)

and consequentlyD(t) = D0 . (102)

Now Eqs. (98) and (99) become

x = −2a−1(at+ 1)(D0 − 1) , (103)y = D0r . (104)

Plugging Eqs. (103) and (104) into Eq. (95), we have

D0 =2

3(105)

Hence,

x =2(1 + at)

3a, y =

2

3r , (106)

ξa =2(1 + at)

3a

(∂

∂t

)a+

2r

3

(∂

∂r

)a. (107)

Thus, we have proven that the HMN spacetime is continuous self-similar only for c = 0 anda 6= 0.

12

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